[leetcode]435. Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]Output: 1Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

Input: [ [1,2], [1,2], [1,2] ]Output: 2Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

Input: [ [1,2], [2,3] ]Output: 0Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

讲真，一点都不想写这个的题解

被自己蠢哭了，告诉我这个题和看电视节目那个贪心入门有什么区别？？？

怎么还能第一关键字是start？？？后一个和前一个比较的是end，当然是第一关键字是end排序啊！！！

大二下开学还是老陈讲的啊………………

这都能写错，还搜题解…………

赶紧老实儿的一天刷一个题吧

/** * Definition for an interval. * struct Interval { *     int start; *     int end; *     Interval() : start(0), end(0) {} *     Interval(int s, int e) : start(s), end(e) {} * }; */bool comp(const Interval &a,const Interval &b)    {        if(a.end==b.end)return a.start<b.start;        return a.end<b.end;    }class Solution {public:        int eraseOverlapIntervals(vector<Interval>& intervals) {        if(intervals.size()==0) return 0;        sort(intervals.begin(),intervals.end(),comp);        int num=1;        int right=intervals[0].end;        for(int i=1;i<intervals.size();i++)        {            if(intervals[i].start<right) continue;            num++;            right=intervals[i].end;        }        return  intervals.size()-num;    }};